GIL(全局解释器锁)
"""
In CPython, the global interpreter lock, or GIL, is a mutex that prevents multiple
native threads from executing Python bytecodes at once. This lock is necessary mainly
because CPython’s memory management is not thread-safe. (However, since the GIL
exists, other features have grown to depend on the guarantees that it enforces.)
"""
"""
GIL是一个互斥锁:保证数据的安全(牺牲效率来换取数据的安全)
阻止同一个进程内,多个线程同时执行(不能并行但能够实现并发)
并发:看起来像同时进行(其实是快速切换并保存当前状态)
GIL全局解释器存在的原因因为Cpython解释器的内存管理不是线程安全的
垃圾回收机制:
1.引用计数
2.标记清除
3.分代回收
同一个进程下的多个线程不能实现并行,但能实现并发,多个线程下的线程能够实现并行。
python多线程是不是没用了?
四个任务:计算密集型任务,每任务个耗时10s。
单核情况:
多线程好一点,消耗的资源少
多核情况:
多线程好一点
多线程和多进程都有自己的优点,根据项目需求合理选择。
"""
# 计算密集型
# from multiprocessing import Process
# from threading import Thread
# import os,time
#
#
# def work():
# res = 0
# for i in range(100):
# res *= i
#
# if __name__ == '__main__':
# l = []
# print(os.cpu_count())
# start = time.time()
# for i in range(4):
# # p = Process(target=work)
# p = Thread(target=work)
# l.append(p)
# p.start()
# for p in l:
# p.join()
# stop = time.time()
# print('run time is %s'%(stop-start))
# 执行结果:
# 进程:
# 4
# run time is 6.02131462097168
# 线程:
# 4
# run time is 0.21412372589111328
# IO密集型
# from multiprocessing import Process
# from threading import Thread
# import os,time
#
#
# def work():
# time.sleep(2)
#
# if __name__ == '__main__':
# l = []
# print(os.cpu_count())
# start = time.time()
# for i in range(100):
# p = Process(target=work) #104.06727409362793
# # p = Thread(target=work) # 2.0227677822113037
# l.append(p)
# p.start()
# for p in l:
# p.join()
# stop = time.time()
# print('run time is %s' %(stop - start))
Gil与普通锁
from threading import Thread,Lock
import time
mutex = Lock()
n = 100
def task():
global n
mutex.acquire()
tmp = n
time.sleep(0.1)
n = tmp - 1
mutex.release()
t_list = []
for i in range(100):
t = Thread(target=task)
t.start()
t_list.append(t)
for t in t_list:
t.join()
print(n)
"""
对于不同的数据,想要保证安全,需要加不同的锁处理
GIL并不能保证数据的安全,他是对Cpython解释器加锁,针对的是线程
保证同一进程下多个线程之间的安全
"""
死锁与递归锁
from threading import Thread,Lock,RLock
import time
"""
自定义锁一次acquire必须对应一次relaese,不能连续acquire
递归锁可以连续的acquire,每acquire一次计数加一:针对第一个抢到我的人
"""
import random
# mutexA = Lock()
# mutexB = Lock()
mutexA = mutexB = RLock() # 抢锁之后会有一个计数,抢一次,计数加一,针对一个抢到我的人
class MyThread(Thread):
def run(self):
self.func1()
self.func2()
def func1(self):
mutexA.acquire()
print('%s 抢到A锁了' % self.name)
mutexB.acquire()
print('%s 抢到B锁了' % self.name)
mutexA.release()
print('%s 释放A锁了' % self.name)
mutexB.release()
print('%s 释放B锁了' % self.name)
def func2(self):
mutexB.acquire()
time.sleep(1)
print('%s 抢到B锁了' % self.name)
mutexA.acquire()
print('%s 抢到A锁了' % self.name)
mutexA.release()
print('%s 释放A锁了' % self.name)
mutexB.release()
print('%s 释放B锁了' % self.name)
for i in range(100):
t = MyThread()
t.start()
event事件
import random
from threading import Event, Thread
import time
event = Event()
def light():
print('红灯亮着!')
time.sleep(3)
event.set() # 解除阻塞,给我的event发了一个信号
print('绿灯亮了!')
def car(i):
print('%s s2q正在等待红灯' % i)
event.wait() # 阻塞
print('%s s2q开始飙车了!'% i)
t1 = Thread(target=light)
t1.start()
for i in range(10):
t = Thread(target=car, args=(i,))
t.start()
信号量
from threading import Thread,Semaphore
import time
import random
sm = Semaphore(5) # 五个厕所五把锁
# 跟普通的互斥锁的区别:
# 普通的互斥锁是独立卫生间,所有人抢一把锁
# 信号量 公共卫生间 有多个坑,所有人抢多把锁
def task(name):
sm.acquire()
print('%s正在上厕所!'% name)
# 模拟上厕所耗时
time.sleep(random.randint(1,5))
sm.release()
if __name__ == '__main__':
for i in range(20):
t= Thread(target=task, args=('sg %s号'% i,))
t.start()
几种不同的queue
import queue
# 1.普通q
# 2.先进后出q
# 3.优先级q
# 先进先出
# q = queue.Queue(3)
# q.put(1)
# q.put(2)
# q.put(3)
# print(q.get())
# print(q.get())
# print(q.get())
# 先进后出
# q = queue.LifoQueue(5)
# q.put(1)
# q.put(2)
# q.put(3)
# q.put(4)
# q.put(5)
# print(q.get())
# 优先级q
q = queue.PriorityQueue()
q.put((10, 'a'))
q.put((-1, 'b'))
q.put((-6, 's2g'))
print(q.get())
print(q.get())
print(q.get())